Let PQR be a right angled isosceles triangle right angled at P (2,1). If equation of line QR is $2x+y=3$, equations of PQ&PR Let PQR be a right angled isosceles triangle right angled at P (2,1). If equation of line QR is $2x+y=3$, find combined equation of pair of lines PQ and PR.
 A: The equation of any straight line passing through $P(2,1)$ can be written as $$\frac{y-1}{x-2}=m$$ where $m$ is the gradient/slope
Now as $\angle RPQ=90^\circ$ and $|PQ|=|RP|$
Using Isosceles Triangles have Two Equal Angles, $\displaystyle \angle PQR=\angle QRP=45^\circ$
If the angle between two lines with slope $m_1,m_2$ is $\phi$
$\displaystyle\tan\phi=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$
In our case, $\displaystyle \phi=45^\circ$
we can set $m_1=m$(to be determined) and the $m_2,$ the gradient of $\displaystyle QR:2x+y=3\iff y=-2x+3$ is $-2$
A: Let $Q=(v,w),\ R=(x,y)$ so that $w=3-2v,\ y=3-2x.$ We want a right angle $QPR$ where $P=(2,1),$ so the dot product of the vectors $PQ=(v-2,w-1)$ and $PR=(x-2,y-1)$ is zero. Replacing $w,y$ in terms of $v,x$ in this dot product, and solving when it is zero for $v$ in terms of $x$ gives
$$v=\frac{6x-8}{5x-6}.$$
Now since the triangle is isosceles, we need equal squared lengths for segments $PQ,PR$ which means
$$(v-2)^2+(w-1)^2-(x-2)^2-(y-1)^2=0, \tag{1}$$
and if we replace things all in terms of $x$ and factor the result (maple is a help here), we get a fraction whose denominator is $(5x-6)^2$ and whose numerator has two linear factors $(5x-8)(5x-4)$ and a third factor which is quadratic irreducible (no real roots). Now we can rule out $x=6/5$ since it leads to $y=3-2(6/5)=3/5$, and then this point lies on the line through the origin and the right angle $P$, so cannot be a solution. This leaves two possible values $4/5,\ 8/5$ for $x$. (That there are two is because in naming the other two points $Q,R$ the names could be switched.)
Thus we now know the points $Q,R$ are in one or the other order $(4/5,7/5),\ (8/5,-1/5).$ It's now routine to get the two equations for lines $PQ,PR$ as 
$$x+3y-5=0,\quad 3x-y=5=0,$$ and if by the phrase of the question the "combined equation" is that which holds when a point is on the union of the two lines, one can form the product of the two left sides here and set it to zero.
